Termination w.r.t. Q of the following Term Rewriting System could not be shown:
Q restricted rewrite system:
The TRS R consists of the following rules:
le2(0, y) -> true
le2(s1(x), 0) -> false
le2(s1(x), s1(y)) -> le2(x, y)
minus2(x, 0) -> x
minus2(0, x) -> 0
minus2(s1(x), s1(y)) -> minus2(x, y)
gcd2(0, y) -> y
gcd2(s1(x), 0) -> s1(x)
gcd2(s1(x), s1(y)) -> if_gcd3(le2(y, x), s1(x), s1(y))
if_gcd3(true, x, y) -> gcd2(minus2(x, y), y)
if_gcd3(false, x, y) -> gcd2(minus2(y, x), x)
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
le2(0, y) -> true
le2(s1(x), 0) -> false
le2(s1(x), s1(y)) -> le2(x, y)
minus2(x, 0) -> x
minus2(0, x) -> 0
minus2(s1(x), s1(y)) -> minus2(x, y)
gcd2(0, y) -> y
gcd2(s1(x), 0) -> s1(x)
gcd2(s1(x), s1(y)) -> if_gcd3(le2(y, x), s1(x), s1(y))
if_gcd3(true, x, y) -> gcd2(minus2(x, y), y)
if_gcd3(false, x, y) -> gcd2(minus2(y, x), x)
Q is empty.
Q DP problem:
The TRS P consists of the following rules:
IF_GCD3(true, x, y) -> GCD2(minus2(x, y), y)
IF_GCD3(true, x, y) -> MINUS2(x, y)
LE2(s1(x), s1(y)) -> LE2(x, y)
IF_GCD3(false, x, y) -> MINUS2(y, x)
GCD2(s1(x), s1(y)) -> LE2(y, x)
MINUS2(s1(x), s1(y)) -> MINUS2(x, y)
GCD2(s1(x), s1(y)) -> IF_GCD3(le2(y, x), s1(x), s1(y))
IF_GCD3(false, x, y) -> GCD2(minus2(y, x), x)
The TRS R consists of the following rules:
le2(0, y) -> true
le2(s1(x), 0) -> false
le2(s1(x), s1(y)) -> le2(x, y)
minus2(x, 0) -> x
minus2(0, x) -> 0
minus2(s1(x), s1(y)) -> minus2(x, y)
gcd2(0, y) -> y
gcd2(s1(x), 0) -> s1(x)
gcd2(s1(x), s1(y)) -> if_gcd3(le2(y, x), s1(x), s1(y))
if_gcd3(true, x, y) -> gcd2(minus2(x, y), y)
if_gcd3(false, x, y) -> gcd2(minus2(y, x), x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
IF_GCD3(true, x, y) -> GCD2(minus2(x, y), y)
IF_GCD3(true, x, y) -> MINUS2(x, y)
LE2(s1(x), s1(y)) -> LE2(x, y)
IF_GCD3(false, x, y) -> MINUS2(y, x)
GCD2(s1(x), s1(y)) -> LE2(y, x)
MINUS2(s1(x), s1(y)) -> MINUS2(x, y)
GCD2(s1(x), s1(y)) -> IF_GCD3(le2(y, x), s1(x), s1(y))
IF_GCD3(false, x, y) -> GCD2(minus2(y, x), x)
The TRS R consists of the following rules:
le2(0, y) -> true
le2(s1(x), 0) -> false
le2(s1(x), s1(y)) -> le2(x, y)
minus2(x, 0) -> x
minus2(0, x) -> 0
minus2(s1(x), s1(y)) -> minus2(x, y)
gcd2(0, y) -> y
gcd2(s1(x), 0) -> s1(x)
gcd2(s1(x), s1(y)) -> if_gcd3(le2(y, x), s1(x), s1(y))
if_gcd3(true, x, y) -> gcd2(minus2(x, y), y)
if_gcd3(false, x, y) -> gcd2(minus2(y, x), x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 3 SCCs with 3 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
MINUS2(s1(x), s1(y)) -> MINUS2(x, y)
The TRS R consists of the following rules:
le2(0, y) -> true
le2(s1(x), 0) -> false
le2(s1(x), s1(y)) -> le2(x, y)
minus2(x, 0) -> x
minus2(0, x) -> 0
minus2(s1(x), s1(y)) -> minus2(x, y)
gcd2(0, y) -> y
gcd2(s1(x), 0) -> s1(x)
gcd2(s1(x), s1(y)) -> if_gcd3(le2(y, x), s1(x), s1(y))
if_gcd3(true, x, y) -> gcd2(minus2(x, y), y)
if_gcd3(false, x, y) -> gcd2(minus2(y, x), x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.
MINUS2(s1(x), s1(y)) -> MINUS2(x, y)
Used argument filtering: MINUS2(x1, x2) = x2
s1(x1) = s1(x1)
Used ordering: Quasi Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
↳ QDP
↳ QDP
Q DP problem:
P is empty.
The TRS R consists of the following rules:
le2(0, y) -> true
le2(s1(x), 0) -> false
le2(s1(x), s1(y)) -> le2(x, y)
minus2(x, 0) -> x
minus2(0, x) -> 0
minus2(s1(x), s1(y)) -> minus2(x, y)
gcd2(0, y) -> y
gcd2(s1(x), 0) -> s1(x)
gcd2(s1(x), s1(y)) -> if_gcd3(le2(y, x), s1(x), s1(y))
if_gcd3(true, x, y) -> gcd2(minus2(x, y), y)
if_gcd3(false, x, y) -> gcd2(minus2(y, x), x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
LE2(s1(x), s1(y)) -> LE2(x, y)
The TRS R consists of the following rules:
le2(0, y) -> true
le2(s1(x), 0) -> false
le2(s1(x), s1(y)) -> le2(x, y)
minus2(x, 0) -> x
minus2(0, x) -> 0
minus2(s1(x), s1(y)) -> minus2(x, y)
gcd2(0, y) -> y
gcd2(s1(x), 0) -> s1(x)
gcd2(s1(x), s1(y)) -> if_gcd3(le2(y, x), s1(x), s1(y))
if_gcd3(true, x, y) -> gcd2(minus2(x, y), y)
if_gcd3(false, x, y) -> gcd2(minus2(y, x), x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.
LE2(s1(x), s1(y)) -> LE2(x, y)
Used argument filtering: LE2(x1, x2) = x2
s1(x1) = s1(x1)
Used ordering: Quasi Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
↳ QDP
Q DP problem:
P is empty.
The TRS R consists of the following rules:
le2(0, y) -> true
le2(s1(x), 0) -> false
le2(s1(x), s1(y)) -> le2(x, y)
minus2(x, 0) -> x
minus2(0, x) -> 0
minus2(s1(x), s1(y)) -> minus2(x, y)
gcd2(0, y) -> y
gcd2(s1(x), 0) -> s1(x)
gcd2(s1(x), s1(y)) -> if_gcd3(le2(y, x), s1(x), s1(y))
if_gcd3(true, x, y) -> gcd2(minus2(x, y), y)
if_gcd3(false, x, y) -> gcd2(minus2(y, x), x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
IF_GCD3(true, x, y) -> GCD2(minus2(x, y), y)
GCD2(s1(x), s1(y)) -> IF_GCD3(le2(y, x), s1(x), s1(y))
IF_GCD3(false, x, y) -> GCD2(minus2(y, x), x)
The TRS R consists of the following rules:
le2(0, y) -> true
le2(s1(x), 0) -> false
le2(s1(x), s1(y)) -> le2(x, y)
minus2(x, 0) -> x
minus2(0, x) -> 0
minus2(s1(x), s1(y)) -> minus2(x, y)
gcd2(0, y) -> y
gcd2(s1(x), 0) -> s1(x)
gcd2(s1(x), s1(y)) -> if_gcd3(le2(y, x), s1(x), s1(y))
if_gcd3(true, x, y) -> gcd2(minus2(x, y), y)
if_gcd3(false, x, y) -> gcd2(minus2(y, x), x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.